What is the integral of 3/(t^4)cos(1/(t^3)-9) ?

The integral of 3/(t^4)cos(1/(t^3)-9) is -sin(1/(t^3)-9)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 3/(t^4)cos(1/(t^3)-9)

\int \frac{3}{t ^{4}} cos (\frac{1}{t ^{3}} - 9) d t

- sin (\frac{1}{t ^{3}} - 9) + C

Solution steps

\int \frac{3}{t ^{4}} cos (\frac{1}{t ^{3}} - 9) d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int \frac{1}{t ^{4}} cos (\frac{1}{t ^{3}} - 9) d t

Apply u-substitution

: \int - \frac{cos ( u )}{3} d u

= 3 \cdot \int - \frac{cos ( u )}{3} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 (- \frac{1}{3} \cdot \int cos (u) d u)

Use the common integral:

\int cos (u) d u = sin (u)

= 3 (- \frac{1}{3} sin (u))

Substitute back

u = \frac{1}{t ^{3}} - 9

= 3 (- \frac{1}{3} sin (\frac{1}{t ^{3}} - 9))

Simplify

3 (- \frac{1}{3} sin (\frac{1}{t ^{3}} - 9)) : - sin (\frac{1}{t ^{3}} - 9)

= - sin (\frac{1}{t ^{3}} - 9)

Add a constant to the solution

= - sin (\frac{1}{t ^{3}} - 9) + C

\int_{- 8}^{8} \frac{1}{2} (64 - x^{2}) d x

n = 1 \sum \infty \frac{8}{n ^{4}}

y^{^{'}} sin (x) + y cos (x) = 0

147.456 π t

\frac{\partial}{\partial x} (x^{2} y^{3} + 2 x y^{- 3} + 5 x + 3 y + 11)

What is the integral of 3/(t^4)cos(1/(t^3)-9) ?
The integral of 3/(t^4)cos(1/(t^3)-9) is -sin(1/(t^3)-9)+C